An astronaut standing on the edge of a cliff on the moon throws a rock directly upward and observes that it passes her on the way down exactly 8 seconds later. 6 Seconds after that, the rock hits the ground at the base of the cliff. Use this information to determine the inital velocity and the height of the cliff.
Note: The acceleration due to the force of gravity on the moon is -5.5 ft/s^2. Therefore, the height of the rock as a function of time, s(t) is given by: s(t)-2.75t^2+v(initial)t+s(initial), where v(initial) denoted the initial velocity and s(initial) denotes the initial height.
Note: The acceleration due to the force of gravity on the moon is -5.5 ft/s^2. Therefore, the height of the rock as a function of time, s(t) is given by: s(t)-2.75t^2+v(initial)t+s(initial), where v(initial) denoted the initial velocity and s(initial) denotes the initial height.